What does $\mathbb{R}∗\mathbb{R}$ mean? I'm sure this has been asked before, but I do not know how to search for notations in past questions.

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I have never seen that before. Possibly the same as $\mathbb{R}\times\mathbb{R}$ as $*$ is sometimes used to denote multiplication in other settings, but that's just a guess. – Alex Becker Dec 28 '11 at 23:28
Could you tell us more about the context where you saw this notation? – Álvaro Lozano-Robledo Dec 28 '11 at 23:29
May be a free product – Norbert Dec 28 '11 at 23:30
I was reading the this section of StackExchange, and saw something that said $f: \mathbb{Z} * \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}$. – UserUsingTheInterwebs Dec 28 '11 at 23:31
Thinking back, I think Norbert is right. I'm pretty sure I've seen this used for either free product or abelian free product. – Alex Becker Dec 28 '11 at 23:33

In group theory, $G ∗ H$ is defined as the free product of $G$ and $H$. It is an operation that constructs a new group which contains both $G$ and $H$ as subgroups, since it is generated by the elements of these groups. For definition, maybe the following link can be of help: http://mathworld.wolfram.com/FreeProduct.html.

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Hatcher uses $X*Y$ to denote the join of two spaces, that is, the quotient of $X \times Y \times I$ by the identifications $(x, y_1, 0)$ with $(x, y_2, 0)$ and $(x_1, y, 1)$ with $(x_2, y, 1)$. This is the space of all line segments joining points of $X$ with points of $Y$. See page 9 of Hatcher's book on algebraic topology for more information.

If $X$ and $Y$ are closed intervals, the cube $X \times Y \times I$ gets collapsed to a tetrahedron. In the case of $X=Y=\mathbb{R}$, I guess we get an "infinite tetrahedron".

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This is not an answer to the question. Since it is not easy to search for mathematical notation, perhaps this can help to confirm where the OP saw the notation. I am making this CW - feel free other occurrences from math.SE which may be relevant.

Searching for "\mathbb Z \ast \mathbb Z" site:math.stackexchange.com lead me to Georges Elencwajg's comment

Dear Akhil, I don't know about the existence of a purely algebraic proof. What I know is that until a few years ago there was no purely algebraic computation of the algebraic fundamental group of the projective complex line minus three points (namely $\pi_1^{alg}(\mathbb P^1_{\mathbb C} \setminus {0,1, \infty})=\widehat{\mathbb Z\ast \mathbb Z}$).

The first is, Is $i_*: \pi_1(S^1 \vee S^1) \to \pi_1(S^1 \times S^1)$ injective? My intuition is that no, this is not injective because $\pi_1(S^1 \vee S^1) = \mathbb{Z} * \mathbb{Z}$, the free group on two generators and $\pi_1 (S^1 \times S^1) = \mathbb{Z}\times \mathbb{Z}$. However, I am not sure if this is in fact true and I am trying to figure out the best way to go about showing it.

And from an answer to the same question:

Consider a homomorphism $f: \mathbb{Z} \ast \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$. If $a$ and $b$ are the generators of $\mathbb{Z} \ast \mathbb{Z}$ then consider what $ab$ and $ba$ map to:
$f(ab) = f(a)f(b) = f(b) f(a) = f(ba)$

In both cases it seems to denote free product.

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